Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int m, n, i, j, z, min;
  cin >> m >> n;
  long long int a[m][n], b[m][n];
  long long int rsum[m], csum[n];
  long long int rc[m], cc[n];
  long long int count = 0;
  long long int cbsum[n];
  long long int rbsum[m];
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      cin >> a[i][j];
      b[i][j] = 0;
    }
  }
  if (n >= m) {
    for (i = 0; i < m; i++) {
      rsum[i] = 0;
      rc[i] = 0;
    }
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        if (j == 0)
          rsum[i] = a[i][j];
        else if (a[i][j] < rsum[i])
          rsum[i] = a[i][j];
      }
    }
    for (i = 0; i < m; i++) {
      long long int z = rsum[i];
      for (j = 0; j < n; j++) {
        b[i][j] = b[i][j] + z;
      }
      count = count + z;
      rc[i] = rc[i] + z;
    }
    for (i = 0; i < n; i++) {
      csum[i] = 0;
      cbsum[i] = 0;
      cc[i] = 0;
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        csum[i] = csum[i] + a[j][i];
        cbsum[i] = cbsum[i] + b[j][i];
      }
      long long int t = csum[i] - cbsum[i];
      if (t >= 0 && t % m == 0) {
        count = count + t / m;
        cc[i] = cc[i] + t / m;
      } else {
        count = -1;
        break;
      }
    }
    cout << count << endl;
    if (count != -1) {
      for (i = 0; i < m; i++) {
        if (rc[i] > 0) {
          for (j = 0; j < rc[i]; j++)
            cout << "row"
                 << " " << i + 1 << endl;
        }
      }
      for (i = 0; i < n; i++) {
        if (cc[i] > 0) {
          for (j = 0; j < cc[i]; j++)
            cout << "col"
                 << " " << i + 1 << endl;
        }
      }
    }
  } else {
    for (i = 0; i < n; i++) {
      csum[i] = 0;
      cc[i] = 0;
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (j == 0)
          csum[i] = a[j][i];
        else if (a[j][i] < csum[i])
          csum[i] = a[j][i];
      }
    }
    for (i = 0; i < n; i++) {
      long long int y = csum[i];
      for (j = 0; j < m; j++) {
        b[j][i] = b[j][i] + y;
      }
      count = count + y;
      cc[i] = cc[i] + y;
    }
    for (i = 0; i < m; i++) {
      rsum[i] = 0;
      rbsum[i] = 0;
      rc[i] = 0;
    }
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        rsum[i] = rsum[i] + a[i][j];
        rbsum[i] = rbsum[i] + b[i][j];
      }
      long long int x = rsum[i] - rbsum[i];
      if (x >= 0 && x % n == 0) {
        count = count + x / n;
        rc[i] = rc[i] + x / n;
      } else {
        count = -1;
        break;
      }
    }
    cout << count << endl;
    if (count != -1) {
      for (i = 0; i < m; i++) {
        if (rc[i] > 0) {
          for (j = 0; j < rc[i]; j++)
            cout << "row"
                 << " " << i + 1 << endl;
        }
      }
      for (i = 0; i < n; i++) {
        if (cc[i] > 0) {
          for (j = 0; j < cc[i]; j++)
            cout << "col"
                 << " " << i + 1 << endl;
        }
      }
    }
  }
}